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KonstantinChe [14]
2 years ago
4

The position of an object at any time t is given by s(t)=3t4−40t3+126t2−9. i. Determine the velocity of the object at any time t

. [5 marks] ii.Does the object ever stop changing? [5 marks] iii.When is the object moving to the right and when is the object moving to the left?
Physics
1 answer:
kozerog [31]2 years ago
0 0

Answer:

1.) V = 12t^3 - 120t^2 + 252t

2.) No

3.) At positive V and negative V

Explanation: Given that the position of an object at any time t is s(t)=3t4−40t3+126t2−9.

1.) To determine the velocity V, we will differenciate the equation above with respect to t.

V = ds/dt

ds/dt = 12t^3 - 120t^2 + 252t

Therefore,

V = 12t^3 - 120t^2 + 252t

2.) The equation above depict an exponential equation which proves that the object never stops changing.

3.) The object moves to the right at positive velocity V and moves to the left at negative velocity V

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An automobile which set the world record for acceleration increase speed from rest to 96 km/h in 3.07 seconds what distance trav
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Answer:

41.02m

Explanation:

Given parameters:

Initial velocity = 0m/s

Final velocity  = 96km/hr

Time taken  = 3.07s

Unknown:

Distance traveled by the time the final speed was achieved = ?

Solution:

To solve this problem, we first find the acceleration of the car;

     Acceleration  = \frac{v - u }{t}

v is the final velocity

u is the initial velocity

t is the time taken

  Now convert the the final velocity to m/s;

          96km/hr to m/s;

               1 km/hr  = 0.278m/s

               96km/hr  = 96 x 0.278 = 26.7m/s

Now;

 Acceleration  = \frac{26.7 - 0}{3.07}   = 8.69m/s²

So;

   v²  = u²  + 2as

v is the final velocity

u is the initial velocity

a is the acceleration

s is the distance

             26.7²  = 0²  + 2 x 8.69 x s

             712.89  = 17.38s

                  s  = 41.02m

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Help me Plz<br>can you solve the upper photo for me​
anygoal [31]

1) The acceleration of the plane is 1.89 times the acceleration of gravity

2) The force acting on the airplane is 2.22\cdot 10^5 N

Explanation:

1)

The acceleration of the plane is given by

a=\frac{v-u}{t}

where

v is the final velocity of the plane

u is the initial velocity

t is the time elapsed

For the airplane in this problem, we have:

u=0 (it starts from rest)

v=200 km/h \cdot \frac{1000 m/km}{3600 s/h}=55.6 m/s is the final velocity

t=3 s is the time

Substituting, we find

a=\frac{55.6-0}{3}=18.5 m/s^2

The acceleration of gravity is

g=9.8 m/s^2

So, the acceleration of the plane relative to g is

\frac{a}{g}=\frac{18.5}{9.8}=1.89

2)

The average force exerted on the plane is given by Newton's second law:

F=ma

where

F is the force

m is the mass of the airplane

a is the acceleration

In this problem, we have

m=12000 kg is the mass of the plane

a=18.5 m/s^2 is the acceleration

Substituting, we find:

F=(12000)(18.5)=2.22\cdot 10^5 N

Learn more about acceleration and forces:

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A light spring obeys Hooke's law. The spring's unstretched length is 34.0 cm. One end of the spring is attached to the top of a
maxonik [38]

Answer:

Explanation:

Hooke's law is represented by thee formula

F = ke where F is force in N and K is the spring constant.

Initial length of the spring = 34cm = 0.34 m

mass of 7.00kg hung

weight = mg = 7 × 9.8 = 68.6 N

Final length of the spring = 44.5 cm = 0.445 m

extension = final length - initial length = 0.445 m - 0.34 m = 0.105 m

a) F = Ke

K = F / e = 68.6 N /0.105 m = 653.33 N/m = 0.653 kN/m

b) F = 150 N

k = 653.33 N/m

F = ke

150 N / 653.33 N/m = e

e = 0.23 m

new length = 0.34 + 0.23= 0.57 m = 57 cm

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